We have to find the square root of `-1+2i` i.e. `\sqrt{-1+2i}`

We will find the square roots of the complex number of the form x+yi , where x and y are real numbers, by the following method:

Let `z^2=(x+yi)^2=-1+2i`

i.e. `(x^2-y^2)+2xyi=-1+2i`

Comparing real and imaginary terms we get,

`x^2-y^2=-1 ------->...

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We have to find the square root of `-1+2i` i.e. `\sqrt{-1+2i}`

We will find the square roots of the complex number of the form x+yi , where x and y are real numbers, by the following method:

Let `z^2=(x+yi)^2=-1+2i`

i.e. `(x^2-y^2)+2xyi=-1+2i`

Comparing real and imaginary terms we get,

`x^2-y^2=-1 -------> (1)`

`2xy=2` implies `xy=1 ------>(2)`

So from (2) we get, y=1/x . Substituting this in (1) we have,

`x^2-\frac{1}{x^2}=-1`

i.e. `x^4+x^2-1=0`

implies `x^2=\frac{-1\pm\sqrt{5}}{2}`

`=0.62, -1.62`

Therefore, `x=\pm\sqrt{0.62}=\pm 0.79` ``

`x^2=-1.62` is discarded since it gives imaginary value.

hence,

When x=0.79, y= 1.27

x=-0.79 , y= -1.27

i.e we have, `\sqrt{-1+2i}=0.79+1.27i or -0.79-1.27i`

`=\pm (0.79+1.27i)`

Hence the square roots of -1+2i are: `\pm` **(0.79+1.27i)**